1 Separation of Variables I
Our goal is to develop a technique to solve IBVPs on the finite interval. This method is called separa-
tion of variables and it reduces the IBVP into a system of ODEs. This approach will give us a series
representation of the solution, that will converge to the true solution of the IBVP.
We first explain how one would do separation of variables in the easiest case when we have a ho-
mogeneous PDE with homogeneous boundary conditions.
1. Assume the solution u(x, t) is of the form u(x, t) = X(x)T (t).
2. Use linearity to write the general solution as an infinite linear combination of general solutions
that satisfy the PDE and boundary conditions:
(a) Spatial Problem: Solve the spatial eigenvalue problem.
(b) Time Problem: Solve the system of homogeneous time ODEs using the eigenvalues from
the spatial problem.
3. Use the initial conditions to solve for the particular solution from the general solution using the
appropriate Fourier coefficients.
1.1 Example Problems
Problem 1.1. (?) Solve
utt − c2 uxx = 0 0 < x < π, t > 0
u|
t=0 = 0 0<x<π
ut |t=0 = x 0<x<π
ux |x=0 = ux |x=π = 0 t > 0.
Solution 1.1. This is a homogeneous PDE with homogeneous Neumann boundary conditions.
Step 1 — Separation of Variables: We look for a separated solution u(x, t) = X(x)T (t) to our IBVP.
Plugging this into the BVP implies
T 00 (t) X 00 (x)
T 00 (t)X(x) − c2 T (t)X 00 (x) = 0 =⇒ 2
= = −λ.
c T (t) X(x)
This gives the following ODEs (see Remark 1 and Remark 2)
X 00 (x) + λX(x) = 0 and T 00 (t) + c2 λT (t) = 0,
with boundary conditions
T (t)X 0 (0) = T (t)X 0 (π) = 0 =⇒ X 0 (0) = X 0 (π) = 0
since we can assume T (t) 6≡ 0 otherwise we will have a trivial solution.
Step 2 — Spatial Problem: We begin by solving the eigenvalue problem
(
−X 00 = λX 0<x<π
X 0 (0) = X 0 (π) = 0.
The solution to the eigenvalue problem (Week 7 Lecture Summary 1.1.2) is
Page 1 of 16
,July 21, 2020 APM346 – Week 9 Justin Ko
Eigenvalues: λn = n2 for n = 0, 1, 2, . . .
Eigenfunctions: Xn = cos(nx) and X0 = 1.
Step 3 — Time Problem: When n = 0, the time problem is
T000 (t) = 0
which has solution
T0 (t) = A0 + B0 t.
The time problem related to the eigenvalues λn for n ≥ 1 is
Tn00 (t) + c2 n2 Tn (t) = 0 for n = 1, 2, . . .
which has solution
Tn (t) = An cos(cnt) + Bn sin(cnt).
Step 4 — General Solution: By the principle of superposition, the general form of our solution is
∞
X ∞
X
u(x, t) = Tn (t)Xn (x) = A0 + B0 t + An cos(cnt) + Bn sin(cnt) cos(nx).
n=1 n=1
Step 5 — Particular Solution: We now use the initial conditions to recover the particular solution by
solving for the constants An and Bn . The initial conditions imply
∞
X
u(x, 0) = 0 =⇒ A0 + An cos(nx) = 0
n=1
and
∞
X
ut (x, 0) = x =⇒ B0 + Bn cn cos(nx) = x.
n=1
Clearly the first initial condition implies that An = 0 for all n ≥ 0. To find the Bn coefficients, we
decompose x into its Fourier cosine series (or equivalently, decomposing |x| into its full Fourier series
on [−π, π])
∞
π X 2((−1)n − 1)
x= + cos(nx)
2 n=1 πn2
and equate coefficients to conclude
π 2((−1)n − 1) 2((−1)n − 1)
B0 = , Bn cn = 2
=⇒ Bn = .
2 πn cπn3
Therefore, our particular solution is
∞
X
u(x, t) = B0 t + Bn sin(cnt) cos(nx)
n=1
π 2((−1)n −1)
where B0 = 2 and Bn = cπn3 .
Remark 1. In general, if
f (x) = g(t)
for all x and t, then there must exist a C such that f (x) = g(t) = C. To see why, we can take the
partial derivatives to conclude that
∂x f (x) = ∂x g(t) = 0 and ∂t g(t) = ∂t f (x) = 0,
so both f and g must be constant by the mean value theorem. Since f (x) = g(t), this implies the
constant must be the same.
Page 2 of 16
, July 21, 2020 APM346 – Week 9 Justin Ko
Remark 2. We will explain how one can rigorously justify the fact that
T 00 (t)X(x) − c2 T (t)X 00 (x) = 0 =⇒ X 00 (x) + λX(x) = 0 and T 00 (t) + c2 λT (t) = 0
for some λ. Since we are interested in non-trivial solutions, there exists a x0 such that X(x0 ) 6= 0.
This implies that
X 00 (x0 )
T 00 (t)X(x0 ) − c2 T (t)X 00 (x0 ) = 0 =⇒ T 00 (t) − c2 T (t) = 0 =⇒ T 00 (t) + c2 λT (t) = 0
X(x0 )
00
where λ1 := − XX(x (x0 )
0)
. Similarly, for a non-trivial solution to exist, there must be a t0 such that
T (t0 ) 6= 0, which will imply
T 00 (t0 )
T 00 (t0 )X(x) − c2 T (t0 )X 00 (x) = 0 =⇒ X 00 (x) − X(x) = 0 =⇒ X 00 (x) + λ2 X(x) = 0
c2 T (t0 )
00
where λ2 := − cT2 T(t(t00)) . To see why λ1 = λ2 , we can evaluate our original equation at (x0 , t0 ) and use
the fact −λ1 X(x0 ) = X 00 (x0 ) and −c2 λ2 T (t0 ) = T 00 (t0 ) to see that
T 00 (t0 )X(x0 ) − c2 T (t0 )X 00 (x0 ) = 0 =⇒ c2 T (t0 )X(x0 )(λ1 − λ2 ) = 0 =⇒ λ1 = λ2
since c2 T (t0 )X(x0 ) 6= 0. A similar procedure can be done to formally justify “dividing by zero” in all
separation of variables problems.
Problem 1.2. (??) Solve
utt − c2 uxx = 0 0 < x < L, t > 0
u|
t=0 = g(x) 0<x<L
ut |t=0 = h(x) 0<x<L
ux |x=0 = u|x=L = 0 t > 0.
Solution 1.2. This is a homogeneous problem with homogeneous boundary conditions.
Step 1 — Separation of Variables: We look for a separated solution u(x, t) = X(x)T (t) to our IBVP.
Plugging this into the BVP implies
T 00 (t) X 00 (x)
T 00 (t)X(x) − kT (t)X 00 (x) = 0 =⇒ 2
= = −λ.
c T (t) X(x)
This gives the following ODEs
X 00 (x) + λX(x) = 0 and T 00 (t) + c2 λT (t) = 0,
with boundary conditions
T (t)X 0 (0) = 0 = T (t)X(L) =⇒ X 0 (0) = X(L) = 0
since we can assume T (t) 6≡ 0 otherwise we will have a trivial solution.
Step 2 — Spatial Problem: We begin by solving the eigenvalue problem
(
−X 00 = λX 0<x<L
0
X (0) = X(L) = 0.
The solution to the eigenvalue problem (Week 7 Lecture Summary 1.1.5) is
Page 3 of 16
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